Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in PYTHON3):

Solution:
def printMatrix(matrix):
    s = [[str(e) for e in row] for row in matrix]
    lens = [max(map(len, col)) for col in zip(*s)]
    fmt = '\t'.join('{{:{}}}'.format(x) for x in lens)
    table = [fmt.format(*row) for row in s]
    print("_________________")
    print('\n'.join(table))
    print("`````````````````")
n,m = map(int,input().split())
mat = []
ans = []
for i in range(n):
    temp = [int(x) for x in input().split()]
    mat.append(temp)
ok = True
for i in range(n):
    for j in range(m):
        if(j == 0):
            # Row
            if(m > n):
                if(mat[i][j]):
                    MIN = mat[i][j]
                    for row in range(m):
                        MIN = min(MIN, mat[i][row])
                    for __ in range(MIN):
                        ans.append(["row", i+1])
                    for row in range(m):
                        mat[i][row]-=MIN
                # Col
                if(mat[i][j]):
                    MIN = mat[i][j]
                    for col in range(n):
                        MIN = min(MIN, mat[col][j])
                    for __ in range(MIN):
                        ans.append(["col", j+1])
                    for col in range(n):
                        mat[col][j]-=MIN
            else:
                # Col
                if(mat[i][j]):
                    MIN = mat[i][j]
                    for col in range(n):
                        MIN = min(MIN, mat[col][j])
                    for __ in range(MIN):
                        ans.append(["col", j+1])
                    for col in range(n):
                        mat[col][j]-=MIN
                if(mat[i][j]):
                    MIN = mat[i][j]
                    for row in range(m):
                        MIN = min(MIN, mat[i][row])
                    for __ in range(MIN):
                        ans.append(["row", i+1])
                    for row in range(m):
                        mat[i][row]-=MIN
            if(mat[i][j]):
                ok = False
                break
        else:
            # Go only on that column
            if(mat[i][j]):
                MIN = mat[i][j]
                for col in range(n):
                    MIN = min(MIN, mat[col][j])
                for __ in range(MIN):
                    ans.append(["col", j+1])
                for col in range(n):
                    mat[col][j]-=MIN
            if(mat[i][j]):
                ok = False
if(ok):
    print(len(ans))
    for i in ans:
        print(i[0], i[1])
else:
    print(-1)